# A linear transform mapping one billinear map to another may not exist (counter-example)


Let $$\phi:V\times W \to E$$, and $$\psi:V\times W \to G$$ be bilinear maps, such that $$N_1(\psi) \subset N_1(\phi)$$ and $$N_2(\psi) \subset N_2$$. W. H. Greub in his Multilinear Algebra book claims (1st ed., ch 1.1, problem 4b), that in this case there is a linear mapping $$T : G \to E$$, such that $$\phi = T\circ\psi$$.

I think, I found a counter example. Assume that $$V,W,E,G$$ are all standard Hilbert spaces over $$\mathbb{R}$$, with the inner product $$\langle \cdot,\cdot \rangle$$ and orthonormal base $$(e_i)_{i=1}$$, probably even of finite dimension. Let $$\psi(v,w) = \langle A(v),w \rangle e_1,$$ and $$\phi(v,w) = B(v)f(w),$$ where $$A : V \to W$$ and $$B : V \to G$$ are linear with $$\ker A \subset \ker B$$ and $$f : W \to \mathbb{R}$$ is a linear functional with $$(\mathrm{im} \; A)^\bot \subset \ker f$$. Thus, $$\ker A = N_1(\psi) \subset N_1(\phi) = \ker B$$ and $$(\im A)^\bot = N_2(\psi) \subset N_2(\phi) = \ker f$$. But in the case $$\rank B > 1$$, it is clear that $$\dim \im \phi > \dim \mathrm{\im \psi} = 1$$ and it is impossible to linearly map a 1d subspace onto a multidimensional one.

To be concrete take $$V,W,G,E = \mathbb{R}^4$$, take $$A$$ to be an orthogonal projection to the subspace $$\Span(e_1,e_2,e_3)$$, and $$B$$ to be an orthogonal projection to $$\Span(e_1,e_2)$$, while $$f(w) = \langle w, e_1 \rangle$$.

This result contradicts the text, so I need some external justification. Am I right or am I wrong?

The statement, as you wrote it, is obviously false: in the particular case where $$V=W$$, $$E=G=K$$ (where $$K$$ is the base field, $$\mathbb{R}$$ if you want), and $$N_1(\phi)=N_1(\psi)=N_2(\psi)=N_2(\psi)=0$$, it would imply that two nondegenerated quadratic forms are always multiple of each other, which is nonsense. If you did not make a mistake when copying the statement, then the book is wrong.